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I am having trouble how to show which is larger.

$\log4\cdot \log3 \quad\text{or}\quad \log4 - \log3$

base is $10$ for every $\log$.

I really appreciate your help

The Integrator
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Hirotaka Nakagame
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3 Answers3

10

On the one hand, $\log 4$ is at least $\frac{1}{2}$, because $4^2=16>10$. Thus the inequality $\log 4 \log 3 \geq \frac{\log 3}{2}$ holds.

On the other hand, $\log 4 - \log 3 = \log (4/3)$, which is less than $\frac{\log 3}{2}$. (Indeed, $(4/3)^2 < 3$ holds, use this to see the claimed inequality).

Putting these together gives $\log 4 \log 3 \geq \frac{\log 3}{2} > \log 4 - \log 3$.

[Meanwhile here is one exercise for the reader: Suppose the base of the logs were 100 or 1000, instead of 10. Would that change things? (Yes it would]

Mike
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Note that: $$\begin{align}\log4\cdot \log3 \quad&\text{or}\quad \log4 - \log3 \iff \\ \log 3^{\log 4} \quad &\text{or} \quad\log \frac43 \iff \\ 3^{\log 4+1} \quad &\text{or} \quad 4 \iff \\ 3^{\log 4+1} \quad >3^{1/2+1} =\sqrt{27}\quad > 5 \quad &> \quad 4.\end{align}$$

farruhota
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    I think your chain would be more clear if you added the intermediate step between $\log 3^{\log 4} \text{or} \log \frac{4}{3}$ and $3^{\log 4 + 1} \text{or} 4$. – NoOneIsHere May 31 '18 at 20:39
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    or get the log 3 to the other side on the first step and factor it out. – WorldSEnder Jun 01 '18 at 01:01
  • @NoOneIsHere, good point, but dropping logs seemed too easy step, so I multiplied both sides by 3 at once. Thank you for commenting. – farruhota Jun 01 '18 at 03:30
  • @WorldSEnder, good point, though it is a little intricate. I tried to get rid of one of the logs as fast as possible. Thank you for commenting. – farruhota Jun 01 '18 at 03:31
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I'm not an expert, but this is what I would do if I want to avoid just plugging it into the calculator.

I'd try to get everything in terms of something similar, so I'm going to go off of $\log_{10}2$.

$\log 4*\log 3$ is basically $\log 2^2* (\text{something slightly larger than $\log 2$})$. Point is, it's $\log 2$ where the power of 2 is something greater than one.

$\log 4-\log 3$ is rewritten $\log(4/3)$, and 4/3 is 2 to the power of something less than one.

I've compared the powers of 2 within the logarithm, and since logs are an increasing function, the one with the bigger power of 2 within is the larger answer. $\log 4*\log 3$ is the larger of the two.

Au101
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Eri
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    Is there any particular reason for all the \mathsfs everywhere? You can simply use \log for log – Au101 May 31 '18 at 21:27
  • @Au101 Nope! Thanks for letting me know - I'm new at this and didn't know you could use \log instead. – Eri May 31 '18 at 21:54